The problem asks us to find the value of the infinite sum $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$.

1. Problem Description

The problem asks us to find the value of the infinite sum $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$.

2. Solution Steps

The given series $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ is related to the polylogarithm function. The polylogarithm function is defined as

$$Li_s(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^s}$$

In our case, we have $s=2$ and $z=x$. Therefore, the given series is the dilogarithm function $Li_2(x)$.

$$Li_2(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2}$$

The dilogarithm function does not have a simple closed-form expression in terms of elementary functions for general values of $x$. However, we can discuss its properties and some special values.

The dilogarithm is defined for $|x| \leq 1$. It converges for $x = 1$, and in this case

$$Li_2(1) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

Also, $Li_2(0) = 0$.

We also know that

$$Li_2(x) = -\int_0^x \frac{\ln(1-t)}{t} dt$$

Unfortunately, there isn't a simple closed-form solution for the general value of the series. Thus, we can only express it as $Li_2(x)$.

3. Final Answer

The sum is equal to the dilogarithm function $Li_2(x)$.

$$\sum_{n=1}^{\infty} \frac{x^n}{n^2} = Li_2(x)$$